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A cricket ball of mass 150 g has an initial velocity u=(3i^+4j^) m s−1 and a final velocity v = -(3i^+4j^) m s−1 after being hit.

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प्रश्न

A cricket ball of mass 150 g has an initial velocity `u = (3hati + 4hatj)` m s−1 and a final velocity `v = - (3hati + 4hatj)` m s−1 after being hit. The change in momentum (final momentum-initial momentum) is (in kg m s1)

पर्याय

  • zero

  • `-(0.45 hati + 0.6 hatj)`

  • `-(0.9 hati + 1.2 hatj)`

  • `-5 (hati + hatj)`

MCQ
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उत्तर

`-(0.9 hati + 1.2 hatj)`

Explanation:

Given, `u = (3hati + 4hatj)` m/s

And `v = - (3hati + 4hatj)` m/s

Mass of the ball = 150 g = 0.15 kg

Δp = Change in momentum

= Final momentum – Initial momentum

= `mv - mu`

= `m(v - u) = (0.15)  [- (3hati + 4hatj) - (3hati + 4hatj)]`

= `(0.15) xx [ - 6hati - 8hatj]`

= `- [0.15 xx 6hati + 0.15 xx 8hatj]`

= `- [0.9 hati + 1.20 hatj]`

Hence, Δp = `-[0.9 hati + 1.2 hatj]`

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पाठ 5: Laws of Motion - Exercises [पृष्ठ ३०]

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एनसीईआरटी एक्झांप्लर Physics [English] Class 11
पाठ 5 Laws of Motion
Exercises | Q 5.3 | पृष्ठ ३०

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